The solution is short but sweet. It turns out it\u2019s easier to think about the probability of losing<\/em> rather than the probability of winning. Let\u2019s define:
\n\\[
\nP(t) = \\text{Probability of losing given we have just flipped the }t^\\text{th}\\text{ Tail}.
\n\\]If we have just flipped our $t^\\text{th}$ Tail, then we\u2019ll win if we flip $t+1$ Heads in a row (probability $\\frac{1}{2^{t+1}}$). Otherwise, we\u2019ll flip our $(t+1)^\\text{st}$ Tail and we\u2019ll have a chance $P(t+1)$ of losing from that point forward. Mathematically,
\n\\[
\nP(t) = \\left(1-\\frac{1}{2^{t+1}}\\right) P(t+1)
\n\\]Our goal is to find the initial probability of winning, which is $1-P(0)$. It\u2019s clear that $\\lim_{t\\to\\infty}P(t) = 1$, because it becomes progressively less likely we\u2019ll win as the number of Tails flipped accrues. Iterating the above recursion, we have<\/p>\n

$\\displaystyle
\n\\text{Probability of winning} = 1-\\prod_{t=1}^\\infty \\left(1-\\frac{1}{2^t}\\right)
\n$<\/span><\/p>\n

This expression occurs often enough that it was given its own name! It\u2019s a special case of the Euler function<\/a> (just one of many things named after Euler), which is itself a special case of the q-Pochhammer symbol<\/a>.<\/p>\n

There is no way to further simplify the infinite product, unfortunately, but it\u2019s rather easy to approximate it. The probability of winning is approximately $0.711211904913\u2026$, so about $71\\%$.<\/p>\n<\/div>\n<\/body>","protected":false},"excerpt":{"rendered":"

This Riddler puzzle involves a particular coin-flipping game. Here is the problem: I flip a coin. If it\u2019s heads, I\u2019ve won the game. If it\u2019s tails, then I have to flip again, now needing to get two heads in a row to win. If, on my second toss, I get another tails instead of a … Continue reading “A coin-flipping game”<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"om_disable_all_campaigns":false,"_monsterinsights_skip_tracking":false,"_monsterinsights_sitenote_active":false,"_monsterinsights_sitenote_note":"","_monsterinsights_sitenote_category":0,"_uf_show_specific_survey":0,"_uf_disable_surveys":false,"footnotes":""},"categories":[7],"tags":[8,15,2],"class_list":["post-2371","post","type-post","status-publish","format-standard","hentry","category-riddler","tag-probability","tag-recursion","tag-riddler"],"aioseo_notices":[],"aioseo_head":"\n\t\t\n\t\n\t\n\t\n\t\n\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t\n\t\t